Question

part i balancing (3 points per equation—30 points total)

1. balanced the following chemical equations with coefficients.

a) sb(s) + cl₂(g) → sbcl₃(s)
b) c₂h₆(g) + o₂(g) → co₂(g) + h₂o(g)
c) fe(s) + hcl(aq) → fecl₃(aq) + h₂(g)
d) al(s) + h₃po₄(aq) → alpo₄(aq) + h₂(g)
e) al₂(so₄)₃(aq) + bacl₂(aq) → baso₄(s) + alcl₃(aq)
f) n₂o₅(g) → no₂(g) + o₂(g)
g) sno₂(s) + 2h₂(g) → sn(s) + h₂o(g)
h) fes(s) + o₂(g) → fe₂o₃(s) + so₂(g)
i) zns(s) + o₂(g) → zno(s) + so₂(g)
j) n₂h₄(g) + n₂o₄(g) → n₂(g) + h₂o(g)

Explanation:

Step1: Balance the elements

For the reaction \(Sb(s)+Cl_2(g)\to SbCl_3(s)\), we first look at the number of \(Cl\) atoms. On the left - hand side, we have \(2\) \(Cl\) atoms (\(Cl_2\)) and on the right - hand side, we have \(3\) \(Cl\) atoms (\(SbCl_3\)). The least common multiple of \(2\) and \(3\) is \(6\). So we put a coefficient of \(3\) in front of \(Cl_2\) and \(2\) in front of \(SbCl_3\): \(\text{Sb}(s)+3\text{Cl}_2(g)\to2\text{SbCl}_3(s)\). Then, to balance the \(Sb\) atoms, we put a coefficient of \(2\) in front of \(Sb\).

Answer:

\(2\text{Sb}(s)+3\text{Cl}_2(g) = 2\text{SbCl}_3(s)\)